AFMC AFMC Solved Paper-2001

  • question_answer
    A drum of radius R full of liquid of density d is rotated at (d rad/s. The increase in pressure at die centre of the drum will be :

    A) \[\frac{{{\omega }^{2}}{{R}^{2}}{{d}^{2}}}{2}\]                  

    B) \[\frac{\omega R{{d}^{2}}}{2}\]

    C) \[\frac{{{\omega }^{2}}Rd}{2}\]                

    D) \[\frac{{{\omega }^{2}}{{R}^{2}}d}{2}\]

    Correct Answer: D

    Solution :

    When an incompressible and  non-viscous liquids flows in stream lined motion from one place to another then at every point of its path the total energy per unit volume is constant. \[P+\frac{1}{2}\rho {{v}^{2}}+\rho gh=\text{constant}\] where P is pressure, \[\rho \]is density, v is velocity and h is height. Also \[v=R\omega \] where R is radius and o is angular velocity. Since, velocity at centre is zero and density \[\rho =d,\]we have \[{{P}_{1}}+\frac{1}{2}\times 0={{P}_{2}}+\frac{1}{2}d{{v}^{2}}\] \[{{P}_{1}}={{P}_{2}}+\frac{1}{2}d{{(R\omega )}^{2}}\] Thus, increase in pressure \[{{P}_{1}}-{{P}_{2}}=\frac{1}{2}d{{R}^{2}}{{\omega }^{2}}\]


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